763654S HYDRODYNAMICS
Solutions 11
Autumn 2011
1. Discharge from a container with a drain pipe
Water ows out of the reservoir down a pipe of cross-sectional area a, see the gure below.
What is the speed of the issuing jet of water? Estimate the time to empty the reservoir.
Describe what happens if there is a small hole in the pipe half way down.
Solution: First of all assume that there is no viscosity between
Area A
uid and the walls in our construction, then Bernoulli equation is
satised well between the upper surface and the outow point of
h
the pipe:
p0
1
p0
1 2
V + g(H + h) +
= v2 + .
2
ρ
2
ρ
From this and the continuity equation of uncompressible uid
Area a
H
V A = va
velocities V and v are solved (by lecture notes)1 :
r
V =a
r
2g(H + h)
A 2 − a2
v=A
2g(H + h)
.
A 2 − a2
We denote the height of the uid surface from the level of out point of the pipe as eh.
Velocity V gives its rate of change, thus we have dierential equation
p
deh
2g
= −V = −a e
h
2
dt
A − a2
r
which has auxiliar form
e − 12
h
deh = −a
r
A2
2g
dt
− a2
having solution with boundary condition eh(0) = H + h as
e
h(t) =
√
H +h−a
r
g
t
2
2(A − a2 )
2
.
(1)
To solve the time to empty the reservoir we have to solve the equation eh(T ) = H meaning
in terms of equation (1) that
r
√
√
H = H +h−a
1 One
can approximate,
or more complicated.
A a → A2 − a2 ≈ A2
2(A2
g
T
− a2 )
but this does not make the solution procedure either easier
which has solution
p
T =
√ 2(A2 − a2 ) √
h+H − H .
√
a g
Fluid is incompressible, va = vhalf point a, meaning that the velocity of the uid is the
same in all points in the pipe. Particularly, at the point y = H/2 velocity is v , thus from
Bernoulli equation between points y = 0 and y = H/2 we get:
H p1
1
p0
1 2
v +g +
= v2 +
2
2
ρ
2
ρ
⇒
p1 = p0 − ρg
H
.
2
Water is not coming out from small hole. Air is trying to compensate the negative pressure
dierence by owing into the pipe from the small aperture.
2. Venturi ow
Air is drawn at volume rate Q along a horizontal pipe through a contraction. The pipe
is connected to a water tank as sketched in the gure below. Estimate the height h for
which water can be sucked into the vertical pipe attached at the constriction.
Solution: The small pipe connecting lower water reserArea B
voir and the contraction is so small that it is resonable to
Area A
assume its eect to the ow neglible. This assumtion includes the fact that the horizontal water ow is not going
Q
into the small connection pipe. From the conservation of
h
the volume (ow), it is deduced that
Q = vb B = va A,
where vb and va are velocities of ow in parts of the construction at area B and area A,
respectively.
In addition to that, from Bernoulli equation (ignoring gravity in the horizontal ow), the
pressure dierence can be solved in those parts:
1
pa
1 2 pb
vb +
= va2 +
2
ρA
2
ρA
⇒ pb − pa = ρ A
va2 − vb2
2
⇒ pb − pa =
ρA Q2 B 2 − A2
.
2
A2 B 2
This pressure dierence is compensated by the sucked water pilar of height h:
ghρW = pb − pa
⇒h=
Q2 ρA B 2 − A2
.
2gρW A2 B 2
Notice the fraction of the air and water densities in the above expression.
3. Train in a tunnel
A train travels at speed 150 km/h in a tunnel. How is the air pressure inside the train
modied compared to the case that the train would be stationary. Assume the dimensions
of the train are width 3 m height 4 m and length 100 m and the corresponding dimensions
of the tunnel are 5 m 7 m and 2 km. (Hint: do all possible simplifying assumptions so
that you still get a non-vanishing result.)
Solution:
Let's rst notice that the Reynolds number is large Re = V a/ν = 2.77 · 107 , where the
velocity of the train is V = 41.67 m/s, kinematic viscosity of air is ν = 1.5 · 10−5 m2 /s and
the characteristic length scale a = 1 m . This implies
that the thickness of the separation
√
layer around the train is of the order of d = L/ Re ≈ 0.02 m where L = 100 m denotes
the length of the train. Thus, it is reasonable to ignore viscosity and eects of the wake
and the separation layer.
Method 1 for the velocity of air v : Let us consider the situation in the rest frame of the
train, where the mass ow of the air at the locations (a) and (b) is
Qb = Ab v 0 ρ.
Qa = Aa V ρ
Above, the cross sectional areas at the locations of (a) and (b), respectively, Aa and
Ab = Aa − At , are expressed with the cross sectional areas of the tunnel Aa = hw and
the train At = ht wt . Brutally assuming that the air is incompressible, the mass ows are
equal and the average velocity of the air around the train is
v0 =
Aa
V.
Aa − At
In the rest frame of the tunnel, the velocity of the air around the train is
v =v−V =
At
= 21.73 m/s.
Aa − At
Method 2 for the velocity of air v : This consideration is done in the rest frame of the
tunnel. Let us assume that the train does not push air in front of it. Meaning that the
air displaced in front oft the moving train ows to the back of the train in the volume
between the tunnel and the train. Based on this and the assumption of incompressible
air, we formulate using the conservation of mass ow that
ρAt V = ρ(Aa − At )v
giving the same result as above.
⇒
v=
At
V,
Aa − At
Now, in the rest frame of the tunnel, we may apply the Bernoulli equation with respect
to the points (a) and (b). Now, at (a) Va = 0 and pa = p0 :
1
pb
1 2 p0
Va +
= v2 +
2
ρ
2
ρ
and solve the pressure dierence
ρA A2t V 2
ρ
pb − p0 = − v 2 = −
= −283 Pa.
2
2(Aa − At )2
The drawdown of the pressure inside of the train is an noticeable phenomena because
it happens rapidly when the train goes into the tunnel. According to the Finnish Meteorological Institute (www.fmi.), the air pressure drops 100 Pa in a 8 m change of
elevation. Thus, 283 Pa drawdown of the air pressure corresponds elevation change of
roughly 22.6 m.
4. Free fall
A falling object in a medium reaches a terminal velocity where the gravitational force and
the drag force of the medium balance each other. Using the attached graph, estimate the
terminal velocities of the following objects in air:
a) a spherical water drop of diameter of 1 mm,
b) a spherical hail of diameter 1 cm,
c) a paratrooper with a parachute diameter 11 m and total mass 160 kg,
d) what would be the velocity of the hail (diameter 1 cm) if the Stokes law were valid?
The drag force has the form F = 21 CD ρAV 2 , and the graph gives the coecient CD as a
function of the Reynolds number NR = DV /ν . Here A is the cross-sectional area of the
object, V its velocity, ρ is the density of the medium, and ν the kinematic viscosity of
the medium. (Hint: Since you do not know the Reynolds number in the beginning, make
rst a simple guess of CD , and then correct that once you have an estimate of NR .
Solution: When a falling object has reached the terminal velocity vt , the gravitational
and drag forces are in balance Fg = Fd . With Fg = ρo Vo g and the drag force Fd =
1
C ρAvt2 , where ρo and Vo denote the density and the volume of the falling object and,
2 D
we get that
s
vt =
2ρo Vo g
.
CD ρA
(2)
Notice, that it is used dierent notation for the terminal velocity to avoid confusion with
symbol for the volume.
The idea is to rst make an educated guess of the Reynolds number and based on that
look the coecient CD from the table and calculate the terminal velocity vt0 . By using
it, one can make a better estimate for the Reynolds number and coecient CD and get
better estimate for vt .
a) Now with d = 1 · 10−3 and ν = 1.5 · 10−6 m2 /s, the Reynolds number
NR = vt
D
= vt · 66.66 s/m.
ν
Thus, based on a guesstimate NR ≈ 102 and A = πd2 /4 and Vo = πd3 /6, the rst
estimate is Cd = 1.0 and vt0 = 3.3 m/s.
Using vt0 = 3.3 m/s, we get that NR = 220 and CD = 0.7 and nally vt ≈ 4 m/s.
b) Here, one needs to know density of ice ρI = 917 kg/m3 . With the exactly same
procedure as above, the estimate for the terminal velocity of the hail is vt = 16 m/s.
With this velocity, the drag coecient is CD = 0.4.
c) Here, one can assume that the parachute is disc-shaped. Noticing that
NR = vt
d
= vt · 7.3 · 105 s/m
ν
and that the value the coecient CD = 1.0 has no dependence on NR with such
a large Reynolds numbers, there is no need for the iteration loop applied above.
Simply,
r
r
vt =
√
2mg
=2 2
CD ρA
mg
≈ 5 m/s.
CD ρπd2
d) The Stokes law says that the drag coecient is inversely proportional to the velocity:
CD =
24
24ν
=
.
NR
Dvt
Using this result in Eq. (2) or in Fg = Fd , the terminal velocity can be solved
straightforwardly in case of a spherical falling hail:
vt =
g fI d2
= 2776 m/s ≈ 2800 m/s
18 fA ν
Eight times the speed of sound! Must be unphysical.